Alg-tool Graphs and Equations

Section 1.2 Graphs and Equations

Subsection 1. Verify a solution

We can always check a solution to an equation by verifying that it makes the equation true.

Subsubsection Examples

Example 1.28.

Verify that \(x=-5\) is a solution of the equation

\begin{equation*} ~~x^2+2x-15=0 \end{equation*}

We show that substituting \(-5\) for \(x\) makes the equation true. When we substitute a negative number for a variable, we should enclose the number in parentheses.

\begin{equation*} \begin{aligned}[t] x^2+2x-15 \amp = (\alert{-5})^2+2(\alert{-5})-15\\ \amp = 25-10-15=0 \end{aligned} \end{equation*}

Because the expression does equal \(0\text{,}\) we see that \(x=-5\) is a solution.

Example 1.29.

Verify that \(x=-3\) is not a solution of the equation

\begin{equation*} ~~\sqrt{2x+10}-3x=8 \end{equation*}

Solution

We show that substituting \(-3\) for \(x\) does not make the equation true.

\begin{equation*} \begin{aligned}[t] \sqrt{2x+10}-3x \amp = \sqrt{2(\alert{-3})+10}-3(\alert{-3})\\ \amp = \sqrt{4}+9=2+9=11 \not= 8 \end{aligned} \end{equation*}

The left side of the equation does not equal 8 when \(x=-3\text{,}\) so \(x=-3\) is not a solution.

Subsubsection Exercises

Checkpoint 1.30.

Decide whether the given value is a solution of the equation.

\begin{equation*} x^3-3x^2-4x+2=10;~~~~x=-2 \end{equation*}

Answer

Yes

Checkpoint 1.31.

Decide whether the given value is a solution of the equation.

\begin{equation*} \sqrt{3x+5}=10+\sqrt{x+7};~~~~x=9 \end{equation*}

Answer

Checkpoint 1.32.

Decide whether the given value is a solution of the equation.

\begin{equation*} \dfrac{2x-1}{x+1}+2=\dfrac{x+1}{x-1};~~~~x=2 \end{equation*}

Answer

Yes

Checkpoint 1.33.

Decide whether the given value is a solution of the equation.

\begin{equation*} 9-4x=5\sqrt{x+3};~~~~x=6 \end{equation*}

Answer

Subsection 2. Solve linear equations and inequalities with parentheses

Strategy for solving linear equations.

Simplify each side of the equation: apply the distributive law, combine like terms.
Use addition and subtraction to get all the variable terms on one side of the equation, and all constsnt terms on the other side.
Divide both sides by the coefficient of the variable.

Subsubsection Examples

Example 1.34.

Solve \(~~3(2a-4) \ge 4-(1-3a)\)

Solution

First, we remove parentheses by applying the distributive law. Then we can combine like terms on each side of the equation.

Note that the minus sign in front of the parentheses on the right side of the equation applies to both terms inside the parentheses.

\begin{align*} 3(2a-4) \amp \ge 4-(1-3a) \amp \amp \blert{\text{Apply the distributive law.}}\\ 6a-12 \amp \ge 4\alert{-}1\alert{+}3a \amp \amp \blert{\text{Simplify the right side.}}\\ 6a-12 \amp \ge 3+3a \amp \amp \blert{\text{Subtract}~ 3a~ \text{from both sides.}}\\ 3a-12 \amp \ge 3 \amp \amp \blert{\text{Add 12 to both sides.}}\\ 3a \amp \ge 15 \amp \amp \blert{\text{Divide both sides by 3.}}\\ a \amp \ge 5 \end{align*}

Example 1.35.

Solve the inequality \(~~25-6x \gt 3x-2(4-x)\)

Solution

We begin by the same way we solve an equation. For this example, we start by removing the parentheses.

\begin{align*} 25-6x \amp \gt 3x-2(4-x) \amp \amp \blert{\text{Apply the distributive law.}}\\ 25-6x \amp \gt 3x-8+2x \amp \amp \blert{\text{Combine like terms.}}\\ 25-6x \amp \gt 5x-8 \amp \amp \blert{\text{Subtract}~ 5x~ \text{from both sides.}}\\ 25-11x \amp \gt -8 \amp \amp \blert{\text{Subtract 25 from both sides.}}\\ -11x \amp \gt -33 \amp \amp \blert{\text{Divide both sides by -11.}}\\ x \amp \alert{\lt} 3 \amp \amp \blert{\text{Don't forget to reverse the inequality symbol.}} \end{align*}

Recall that if we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality symbol.

Subsubsection Exercises

Checkpoint 1.36.

Solve the inequality \(~~-4(x+2)+3(x-2) \ge -2\)

Answer

\(x \le -12\)

Checkpoint 1.37.

Solve the equation \(~~4(2-3w)=9-3(2w-1)\)

Answer

\(\dfrac{-1}{2}\)

Checkpoint 1.38.

Solve the inequality \(~~2(3h-6) \lt 5-(h-4)\)

Answer

\(h \lt 3\)

Checkpoint 1.39.

Solve the equation \(~~0.25(x+3)-0.45(x-3)=0.30\)

Answer

\(9\)

Subsection 3. Solve an equation in two variables

A solution of an equation in two variables \(x\) and \(y\) is written as an ordered pair, \((x,y)\text{.}\) For example, the solution \((-2,5)\) means that \(x=-2\) and \(y=5\text{.}\)

Subsubsection Examples

Example 1.40.

Is \((-3,2)\) a solution of the equation \(~~x^2+4y^2=25\) ?

Solution

We substitute \(x=\alert{-3}\) and \(y=\alert{2}\) into the equation.

\begin{equation*} (\alert{-3})^2+4(\alert{2})^2=9+4(4)=9+16=25 \end{equation*}

The ordered pair \((-3,2)\) satisfies the equation, so it is a solution.

Example 1.41.

Which of the following ordered pairs are solutions of the equation whose graph is shown?

\(\displaystyle (-3,-2)\)
\(\displaystyle (-5,0)\)
\(\displaystyle (1,-4)\)
\(\displaystyle (-1,-6)\)

Solution

The graph of an equation is just a picture of its solutions, so points that lie on the graph are solutions of the equation.

The points \((-3,-2)\) and \((-1,-6)\) lie on the graph, so they represent solutions of the equation. The points \((-5,0)\) and \((1,-4)\) do not lie on the graph, so they are not solutions of the equation.